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It can be easily seen that there exists a functor $F:Top \rightarrow Grpd$ from the category of topological spaces to the category of groupoids defined as follows:

Obj: $X \mapsto \pi_{\leq 1}(X)$, where $\pi_{\leq 1}(X)$ is the fundamental groupoid of $X$.

Mor: ($f:X \rightarrow Y) \mapsto F(f):\pi_{\leq 1}(X) \rightarrow \pi_{\leq 1}(Y)$ where the functor $F(f)$ is defined as follows:

Obj: $x \mapsto f(x)$

Mor: $([\gamma]:x \rightarrow y) \mapsto [f(\gamma)]:f(x) \rightarrow f(y) $ where $[\gamma]$ is the homotopy class of path $\gamma$ in $X$ and $[f(\gamma)]$ is the homotopy class of path $f (\gamma)$ in $Y$.

Also it is not difficult to see that $F$ is well behaved with homotopy (for example in the chapter 6 of http://www.groupoids.org.uk/pdffiles/topgrpds-e.pdf)) that is in the sense that if $f,g: X \rightarrow Y$ are homotopic then the induced functors $F(f)$ and $F(g)$ are naturally isomorphic.

Also using this functor $F$ one can construct a 2-funntor $\tilde{F}: 1Type \mapsto Gpd$ where $1Type$ is the 2-category consisting of homotopy 1-types, maps and homotopy class of homotopies between maps and $Gpd$ is the 2-category consist of Groupoids, functors and natural transformations. Now according to Homotopy hypothesis of dimension 1 as mentioned in http://math.ucr.edu/home/baez/homotopy/homotopy.pdf this $\tilde{F}$ is an equivalence of 2-categories.

So from the above mentioned observations I felt that the functor $F$ is an interesting object of study.

Now if we consider the following sequence of functors: $$ X \stackrel{F}{\mapsto} \pi_{\le 1}(X) \stackrel{N}{\mapsto} N(\pi_{\le 1}(X)) \stackrel{r}{\mapsto} r(N(\pi_{\le 1}(X))) $$ where $N$ is the nerve functor and $r$ is the geometric realization functor.

My question is the following:

How the topological spaces $X$ and $r(N(\pi_{\leq 1}(X)))$ are related? It may be possible that my question does not make much sense when $X$ is a general topological space but then, does there exist any specific class of topological spaces $X$ which has a "good relation" with $r(N(\pi_{\leq 1}(X)))$?

I would be also very grateful if someone can refer some literature in this direction.

Thank you!

David Roberts
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Adittya Chaudhuri
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    https://en.wikipedia.org/wiki/Postnikov_system – Denis Nardin May 12 '20 at 11:38
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    @DenisNardin Thanks for the comment. But can you please explain in little details how this helps in answering my question. Prior apologies if I sound stupid! – Adittya Chaudhuri May 12 '20 at 11:42
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    @AdittyaChaudhuri $|N \pi X|$ (using more usual notations...) is the $X_1$ from the Wikipedia article. It's a space satisfying $\pi_k(X_1, x) = 0$ for all $x \in X_1$ and $k \ge 2$, equipped with a map $X \to X_1$ inducing a bijection on $\pi_0$ and an isomorphism on all $\pi_1$'s. – Najib Idrissi May 12 '20 at 12:49
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    PS: Note that this is essentially the statement of the dimension 1 homotopy hypothesis you mention (not exactly the same strictly speaking, but I think they're pretty much equivalent). – Najib Idrissi May 12 '20 at 12:59
  • @NajibIdrissi Thank you for the valuable comment. I am trying to understand it. I did not know that this is equivalent to Homotopy Hypothesis in one dimension. This question just came to me out of only curiosity! – Adittya Chaudhuri May 12 '20 at 13:07
  • As far as I know, geometric realization of a simplicial set has structure of a CW complex... May be you can ask if you assume $X$ to be a CW complex, do you get an isomorphism of CW complexes between $X$ and $r(N(\pi_1(X)))$? – Praphulla Koushik May 12 '20 at 16:12
  • @PraphullaKoushik Yes that is true. But I did not mention about isomorphism anywhere in the question. But in this case I already got my answer from Najib Idrissi's comment. – Adittya Chaudhuri May 12 '20 at 17:41
  • So, you do know that there is an isomorphism when X is a CW complex? – Praphulla Koushik May 12 '20 at 18:01
  • @PraphullaKoushik No according to Najib Idrissi's comment it is $X_1$ coming from Postnikov system of $X$. But I am yet to find out how. I am trying. – Adittya Chaudhuri May 12 '20 at 18:16
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    @PraphullaKoushik $X$ and $r(N(\pi_{\le 1}(X)))$ are not isomorphic even when $X$ is a CW-complex. Consider the case of $S^2$. Every loop is trivial, and $\pi_{\le 1}(X)$ is equivalent to the trivial category. So its geometric realization is contractible. – Najib Idrissi May 12 '20 at 20:33
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    @AdittyaChaudhuri No problem at all. Could you say what is not so clear? I would suggest the following steps: show that its path components are the path components of $X$ so that you can restrict yourself to a path connected $X$; show that the higher homotopy groups of $|N \pi X|$ vanish (there is a standard construction to build a contractible covering space, it is somewhere in Hatcher, look for "classifying space"); and show that the map $X \to |N \pi X|$ (obtained by playing with adjunctions) is an isomorphism on $\pi_1$. (This can also all be deduced from Ronnie Brown's answer below.) – Najib Idrissi May 12 '20 at 20:36
  • @NajibIdrissi Thank you very much for the comment. I got what you said. I will check the details rigorously. According to your comment I guess if in particular $X$ is a homotopy 1-type then then the map $X \rightarrow |N\piX|$ is a weak homotopy equivalence . Now I feel that if $X$ is an $n-$ type and instead of fundamental groupoid functor if we start with some "appropriately generalised" n-functor then we will get a weak equivalence from $X$ to its " appropriately generalised" geometric realisation. (appropriately generalised as "the usual realisation" is a functor , not a n-functor. ) – Adittya Chaudhuri May 13 '20 at 02:50
  • @NajibIdrissi I also guess if we move up the ladder i.e if $X$ is an arbitrary $C W$ complex then if we start with "appropriately generalised notion of infinity functor" instead of fundamental groupoid functor then we will get a weak equivalence from $X$ to it's appropriately generalised notion of " geometric realisation " of the "appropriately generalised nerve" of the image of $X$ under that "infinity functor". "I think this is what you meant to say about it's relation with Homotopy Hypothesis". Please let me know if my guess is wrong. I have very little background in these topics. – Adittya Chaudhuri May 13 '20 at 03:03
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    @AdittyaChaudhuri If you look at my answer, you generally do not have a canonical map $X\to \lvert N (\Pi_1) \rvert$. If $X$ is a CW-complex, you can use Whitehead's theorem to construct a homotopy inverse to the map $\lvert \operatorname{Sing}(X)\rvert \xrightarrow{\simeq} X$ then compose with the natural map on the left from my answer, but such a homotopy inverse is not uniquely defined (it is defined only up to a contractible ambiguity). If $X$ is not a CW-complex, such a map might not even exist (take a finite topological model of $S^1$, for example).. – Harry Gindi May 13 '20 at 14:50
  • @HarryGindi So are you trying to say that $\lvert N(\prod_1) \rvert$ is not exactly $X_1$ in the Najib Idrissi's comment because of that "contractible ambiguity" you mentioned in your comment above? – Adittya Chaudhuri May 14 '20 at 06:03
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    @AdittyaChaudhuri I'm sorry but you have to ask Najib what he meant. It's not the case that you can always build a literally continuous map (my example of the finite model for S^1 is the counterexample). – Harry Gindi May 14 '20 at 07:40
  • @HarryGindi But if $X$ is a CW complex then will $\lvert N(\pi_{\leq X}(X)) \rvert$ be always $X_1$? – Adittya Chaudhuri May 14 '20 at 07:50
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    What do you mean X_1? Do you mean the first postnikov stage? Then yes. If you mean the 1-skeleton of X, the answer is no. – Harry Gindi May 14 '20 at 07:53
  • @HarryGindi I meant first postnikov stage. Thanks – Adittya Chaudhuri May 14 '20 at 07:55

3 Answers3

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The inclusion of groupoids into simplicial sets is fully faithful. Its left adjoint, $\Pi_1$ is given by left Kan extension of the functor $\Delta\to \mathcal{Gpd}$ sending the n-simplex to the contractible groupoid with objects $\{0,...,n\}$.

The entirety of the data of the homotopy type of the space $X$ is contained in its singular simplicial set, which is canonically a Kan complex. In particular, the fundamental groupoid functor you've written above is canonically isomorphic to the composite $\Pi_1 \circ \operatorname{Sing}$. Then we have a universal natural transformation $$\operatorname{Sing} \to N\circ \Pi_1\circ \operatorname{Sing}$$ given by the unit of the adjunction $Π_1\dashv N$. Taking geometric realizations, we obtain a span

$$\lvert \Pi_1 \rvert \cong \lvert N\circ \Pi_1\circ \operatorname{Sing}\rvert \leftarrow \lvert \operatorname{Sing}\rvert \xrightarrow{\simeq} \operatorname{id_{\mathbf{Top}}},$$

where the righthand map is the counit of the adjunction between simplicial sets and topological spaces $\lvert \bullet \rvert \dashv \operatorname{Sing}$, and is a natural weak homotopy equivalence by a theorem of Quillen.

So the lefthand map exhibits the nerve of the fundamental groupoid as stage 1 of the Postnikov System as mentioned by Denis in the first comment.

Harry Gindi
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  • Thanks for the answer. Here in the first line by "inclusion of groupoids into simplicial sets" do you mean to say nerve functor? Also you mentioned "Universal natural transformation"... I came across that term in Mac lane (Working Mathematician -Pg 39 2nd edition).. did you mean to say in that way? But in that case a universal natural transformation is defined between 2 functors $T_0,T_1: C \rightarrow C \times 2$ where $2$ is the category $0 \rightarrow 1$ and $C$ is any category. I am not getting how we use this notion here. Also are we using the fact it is Universal? – Adittya Chaudhuri May 13 '20 at 03:52
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    1.) Yes, I mean along the nerve. 2.) By universal natural transformation, I mean 'Universal Morphism' as in this wikipedia article https://en.wikipedia.org/wiki/Adjoint_functors . The unit and counit of an adjunction are always universal in this sense. – Harry Gindi May 13 '20 at 11:23
  • Thanks for the explanation. – Adittya Chaudhuri May 13 '20 at 13:23
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    @AdittyaChaudhuri I just wanted to mention: This is a general pattern. At least when we have a Kan complex K, we can write down its map to its n+2-coskeleton, which is also a Kan complex.Then the unit map for the coskeleton adjunction, $K\to \operatorname{Cosk}{n+2} K$ exhibits $\operatorname{Cosk}{n+2}$ as the nth truncation, that is, the nth stage of the postnikov system. But note that when K is a Kan complex, the $1+2=3$-coskeleton for is canonically isomorphic to $NΠ_1(K)$.Taking the case $K=\operatorname{Sing}(X)$ gives the direct relation to topology by the same span as above. – Harry Gindi May 13 '20 at 14:42
  • Thank you very much for this remark on this "general pattern". I am not much well versed with these notions. I will take some time to completely understand what you are trying to say in the above comment. – Adittya Chaudhuri May 14 '20 at 05:38
  • In the answer you took geometric realisation of $\prod_1$ but $\prod_1$ is category valued . So did you mean to say $\lvert N \circ \prod_1 \rvert$ instead of $\lvert \prod_1 \rvert$ ? – Adittya Chaudhuri May 14 '20 at 05:53
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    yes. that's right. – Harry Gindi May 14 '20 at 07:38
  • Just for confirmation! In that case $\lvert N \circ \pi_1 \rvert$ will be same as $\lvert N \circ \pi_1 \circ Sing \rvert$? Is it? – Adittya Chaudhuri May 14 '20 at 07:47
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    @AdittyaChaudhuri yeah, implicitly they are two different constructions that are canonically isomorphic. – Harry Gindi May 14 '20 at 07:52
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I have now (May 13) partitioned the answer into the blocks 1,2, as I think 2 is the simpler answer!

1 I hope the book Nonabelian Algebraic Topology will answer the question for you.

A groupoid is level one of a structure called a crossed complex which is a kind of nonabelian chain complex but also with the groupoid structure in dimensions $\leqslant 1$, which operates on the higher dimensional stuff. There is a homotopically defined functor $\Pi$ from the category of filtered spaces to crossed complexes, using the fundamental groupoid and relative homotopy groups and also a functor $\mathbb B$ from crossed complexes to filtered spaces such that $\Pi \mathbb B$ is naturally equivalent to the identity. This setup is particularly useful for CW-complexes with their standard cellular filtration.

Part of the thesis of the book is to use structured spaces, in this case filtered spaces, to get to link various dimensions, and in this way to use strict algebraic structures. Also the proofs use higher cubical homotopy groupoids, and are non trivial, but can involve the intuitive idea of allowing "algebraic inverses to subdivision", that is generalising to dimension $n$ the usual composition of paths. This is more difficult to do simplicially.

Part I of the book deals with dimensions $0,1,2$ where it is easier to explain the intuitions, and history. Section 2.4 discusses the classifying space of a group and of a crossed module, but the groupoid case comes in Chapter 11.

2 But an answer can easily be put: a groupoid $G$ has a set of objects say $G_0$ and its classifying space $BG $ also contains the set $G_0$.The fundamental groupoid $\pi_1(BG, G_0)$ is naturally isomorphic to $G$! That is, you need the concept of the fundamental groupoid $\pi_1(X,S) $ on a set $S$ of base points, which is formed of homotopy classes rel end points of paths in $X$ with end points in $S$. You can find this developed in the book "Topology and Groupoids". The notion itself was published in my paper

``Groupoids and Van Kampen's theorem'', Proc. London Math. Soc. (3) 17 (1967) 385-40.

The use of this Van Kampen Theorem involving a set of base points was to allow a theorem which could compute fundamental groups of spaces, such as the circle, where the traditional theorem did not apply.

See also this mathoverflow link.

Ronnie Brown
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  • Thank you Sir very much. I will definitely read the book you mentioned & will try to connect it to answer my question. But geometrically what is the meaning of geometrical realisation of the nerve of a fundamental groupoid? . So according to Najib Idrissi's comment it is $X_1$ which is coming from postkinov system of $X$ i.e sequence of spaces$X_n$ and maps $\phi_n:X \rightarrow X_n$ which induce isomorphisms from $\pi_i(X)$ to $ \pi_i(X_n)$ for $i \leq n$, bijection on $\pi_0$ and $\pi(X_n)=0$ for $i > n$. But how this is related to geometric realisation I am not getting properly! – Adittya Chaudhuri May 12 '20 at 18:10
  • Sir, in the comment above I assumed $X$ is a C W complex and "by sequence of spaces" I mean an inverse system of spaces. – Adittya Chaudhuri May 12 '20 at 18:13
  • Thank you Sir for the edit. I am trying to understand it. – Adittya Chaudhuri May 12 '20 at 20:20
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A very rough argument that can be (easily) formalized is as follows:

We have a notion of $\infty$-groupoids. These are like groupoids, but they have homotopies between morphisms, homotopies between homotopies, and so on. Every topological space presents an infinity groupoid by taking the objects to be points, morphisms to be paths, morphisms between morphisms to be homotopies of paths, etc.

If one takes the connected components of this we get the path components of our space. If one takes the connected components of the automorphism group of a point, we get the fundamental group. If one takes the connected components of the morphisms from a constant path to itself we get the second homotopy group, and so on.

Hence this $\infty$-groupoid can be seen as presenting all the homotopical information of our space. Now the fundamental groupoid is given by taking this $\infty$-groupoid and taking connected components of the morphisms between points to get an actual groupoid. Now we just remarked that the higher homotopy information (homotopy groups after the first) are all contained in the connected components of the morphism sets. By discretizing these sets we are removing all higher homotopical information.

So what should we expect when we realize it? Well we should expect that $\pi_0 , \pi_1$ are that of are spaces, but $\pi_n$ for $n>1$ is trivial. This is exactly what happens, we get the first Postnikov space for $X$, i.e. $K(\pi_1(X),1)$ (or really a disjoint union of these for each path component).

Connor Malin
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  • Thanks for the answer. Can you please explain in little detail about what do you mean by Connected components of Morphisms? Also I did not get what is meant by "discretizing the morphism sets"? Did you mean to say all higher $k-$ morphisms are identity? – Adittya Chaudhuri May 13 '20 at 04:26
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    Connected components of a groupoid mean the isomorphism classes. You are correct in your guess of what "discretizing" meant. – Connor Malin May 13 '20 at 15:55